document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=65100>Question 65100</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do you do partial fraction decomposition?? Like on a problem such as:     1/(x^2+x)???  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #46585 by </B><A HREF=/tutors/aboutme.mpl?userid=FadelSalem><B>FadelSalem(12)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=FadelSalem><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=FadelSalem><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=46585\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hey, the answer is very easy as follwos:\r<BR>\n" );
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document.write( "1/(x^2+x) = 1/ x(x+1) = A/x + B/(x+1)\r<BR>\n" );
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document.write( "Multiply each side by x(x+1), we'll get:\r<BR>\n" );
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document.write( "1 = A(x+1) + B x<BR>\n" );
document.write( "1 = Ax + A + Bx<BR>\n" );
document.write( "Compare absolute terms we get: 1 = A<BR>\n" );
document.write( "compare x coefficients we get: 0 = A + B [and that means B = -1]<BR>\n" );
document.write( "So, 1/(x^2+x) = 1/x - 1/(x+1)\r<BR>\n" );
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document.write( "you're welcome any time and wish you Happy new year<BR>\n" );
document.write( "Fadel Salem - Egypt<BR>\n" );
document.write( "FadelSalem@hotmail.com </SPAN>\n" );
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